What's the difference between Series.sort() and Series.order()?

Question

s = pd.Series( nr.randint( 0, 10, 5 ), index=nr.randint(0, 10, 5 ) )
s

Output

1    3
7    6
2    0
9    7
1    6

Answer 1

Your Question: Is this (Series.sort in-place v.s. Series.order return-new-obj) the only difference between the two methods?

BEFORE pandas 0.17.0 Final release (i.e. before 2015-10-09)

Short Answer: YES. They are functionally equivalent.

Longer Answer:

pandas.Series.sort(): change the object itself (in-place sorting), but returns nothing.

Sort values and index labels by value. This is an inplace sort by default. Series.order is the equivalent but returns a new Series.

So

>>> s = pd.Series([3,4,0,3]).sort()
>>> s

outputs nothing. See the answer here for more details.

pandas.Series.order(): dose not change the object, instead it returns a new sorted object.

Sorts Series object, by value, maintaining index-value link. This will return a new Series by default. Series.sort is the equivalent but as an inplace method.

---

AFTER pandas 0.17.0 Final release (i.e. after 2015-10-09)

The API of sorting is changed, things became cleaner and more pleasant.

To sort by the values, both Series.sort() and Series.order() are DEPRECATED, replaced by the new Series.sort_values() api, which returns a sorted Series object.

To summary the changes (excerpt from pandas 0.17.0 doc):

To sort by the values (A * marks items that will show a FutureWarning):

        Previous              |         Replacement
------------------------------|-----------------------------------
* Series.order()              |  Series.sort_values()
* Series.sort()               |  Series.sort_values(inplace=True)
* DataFrame.sort(columns=...) |  DataFrame.sort_values(by=...)